2018 AMC 12B Problems/Problem 7
Problem
What is the value of ![\[\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?\]](http://latex.artofproblemsolving.com/1/0/8/108141fcccb63d051a626897ded91e10882ab0db.png)
Solution 1
From the Change of Base Formula, we have ![\[\frac{\prod_{i=3}^{13} \log (2i+1)}{\prod_{i=1}^{11}\log (2i+1)} = \frac{\log 25 \cdot \log 27}{\log 3 \cdot \log 5} = \frac{(2\log 5)\cdot(3\log 3)}{\log 3 \cdot \log 5} = \boxed{\textbf{(C) } 6}.\]](http://latex.artofproblemsolving.com/3/d/0/3d0aba10cc169dfcc3b353ffe370b5a527574723.png)
Solution 2
Using the chain rule of logarithms 
we get
Solution 3
Using the Change of Base Formula, we have \frac{\log(7) \cdot \log(9)\cdot \log(11) ... \log(25) \cdot \log(27)}{\log(3) \cdot \log(5)\cdot \log(7)...\log(21) \cdot \log(23)}
From this, we realize that the top logs from \log(7) - \log(23) cancel with the bottom logs from \log(7) - \log(23). This leaves us with \frac{\log(25) \cdot \log(27)}{\log(3) \cdot \log(5). Going from this step, we realize that we can use the communicative property to achieve \frac{\log(25) \cdot \log(27)}{\log(5) \cdot \log(3).
Therefore, we can use the Change of Base Formula to get back to \log_525 \cdot log_327. This equals to 2 \cdot 5 = \boxed{\textbf{(C) } 6}
Video Solution
https://youtu.be/RdIIEhsbZKw?t=605
~ pi_is_3.14
See Also
| 2018 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
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